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Path: news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail
From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: Turing Machine computable functions MUST apply finite string
transformations to inputs
Date: Wed, 30 Apr 2025 23:55:01 -0400
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On 4/30/2025 11:50 PM, olcott wrote:
> On 4/30/2025 7:17 PM, André G. Isaak wrote:
>> On 2025-04-30 16:09, olcott wrote:
>>> On 4/30/2025 2:55 PM, dbush wrote:
>>>> On 4/30/2025 1:32 PM, olcott wrote:
>>>>> On 4/30/2025 11:11 AM, Richard Heathfield wrote:
>>>>>> On 30/04/2025 16:44, joes wrote:
>>>>>>> Am Wed, 30 Apr 2025 10:09:45 -0500 schrieb olcott:
>>>>>>>> On 4/29/2025 5:01 AM, Mikko wrote:
>>>>>>>
>>>>>>>>> Irrelevant. There is sufficient agreement what Turing machines
>>>>>>>>> are.
>>>>>>>>
>>>>>>>> Turing machine computable functions must apply finite string
>>>>>>>> transformation rues to inputs to derive outputs.
>>>>>>>>
>>>>>>>> This is not a function that computes the sum(3,2):
>>>>>>>> int sum(int x, int y) { return 5; }
>>>>>>> Yes it is, for all inputs.
>>>>>>
>>>>>> Not much of a computation, though, is it?
>>>>>>
>>>>>
>>>>> It IS NOT a Turing Computable function
>>>>
>>>> Lying by misuse of terms.
>>>>
>>>> A turing computable function is a mapping for which an algorithm
>>>> exists to compute it, not the algorithm itself.
>>>>
>>>> Further use of "turing computable function" when what is meant is
>>>> "algorithm" will result in the former being replaced with the later
>>>> in future responses to your posts to make it clear what you are
>>>> actually talking about.
>>>>
>>>>
>>>>> because it does not ever apply any finite
>>>>> string transformation rules to its inputs.
>>>>
>>>> Sure it does. It computes the mapping of all pairs of integers to
>>>> the number 5.
>>>>
>>>
>>> int sum(int x, int y) { return 5; }
>>> Does not apply transformations to its inputs
>>> to derive its outputs thus is no kind of computable
>>> function not even for sum(2,3).
>>
>> You are still hopelessly confused about your terminology.
>>
>> Computable functions are a subset of mathematical functions, and
>> mathematical functions are *not* the same thing as C functions.
>> Functions do not apply "transformations". They are simply mappings,
>> and a functions which maps every pair of natural numbers to 5 is a
>> perfectly legitimate, albeit not very interesting, function.
>>
>> What makes this function a *computable function* is that fact that it
>> is possible to construct a C function (or a Turing Machine, or some
>> other type of algorithm) such as int foo(int x, int y) {return 5;}
>> which computes that particular function; but the C function and the
>> computable function it computes are entirely separate entities.
>
> computes the sum of two integers
> by transforming the inputs into an output.
> int sum(int x, int y) { return x + y; }
>
> Computes no function because it ignores its inputs.
> int sum(int x, int y) { return 5; }
False. It computes this function:
For all integers X and Y:
(X,Y) maps to 5
>
>> [I won't call that function 'sum()' because that would be misleading,
>> but the the *name* assigned to a C function has no necessary relation
>> to the function it computes. It's good programming practice to give
>> functions descriptive names but nothing in the C standard requires it).
>>
>> You keep conflating C functions/Turing Machines with computable
>> functions and as a result come across as completely ignorant about the
>> topic you purport to be discussing. No C function or Turing Machine is
>> a computable function. They are ways of expressing algorithms.
>>
>> André
>>
>
> The complete ignorance is to expect HHH(DD)
> to report on DD(DD).
It is expected when we start with the assumption that the following
requirements can be satisfied and HHH satisfies them.
Given any algorithm (i.e. a fixed immutable sequence of instructions) X
described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the
following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly
When we reach a contradiction, that proves the assumption that the above
requirements can be met is false, as shown by Linz and as you have
*explicitly* agreed with.